$\begingroup$This is the elegant solution to this problem. I thought I'd post this exact solution on here if everyone did it the long way, but I'm half a year late. (+1)$\endgroup$
– user70962May 20 '13 at 21:52

$\begingroup$@Dole, 1st equality: addition of an inverse, 2nd equality: formula for inverse of a product, 3rd equality: removal of inverses. Remember in this group, we can add or remove $^{-1}$ from anything, because every element is its own inverse. Does that answer your question?$\endgroup$
– asmeurerSep 14 '17 at 23:27

Proof: let for all $a,b$ in group $G$.
claim that To show $ab=ba$ a commutative.
By using a fact that $a\cdot a=b\cdot b=(ab)\cdot(ab)=e$.
since $(ab)^2=a^2\cdot b^2=e\cdot e=e$. We have $ab\cdot ab=e$. Multiplying on the right by $ba$, we obtain
\begin{align}
ab\cdot ab\cdot ba &= e\cdot ba\\
ab\cdot a(b\cdot b)\cdot a &= ba\\
ab\cdot a\cdot b^2\cdot a &=\\
ab\cdot a\cdot e\cdot a &=\\
ab\cdot a\cdot a &=\\
ab\cdot e &=\\
ab &= ba,
\end{align}
for all $a,b$ in $G$. since $G$ is abelian group. This is proved last.

$\begingroup$I think you meant "therefore" instead of "since" in the last sentence. That $G$ is abelian was to be proved, not given (and BTW, nowhere you've used that it is specifically a group; the argumentation works just as well for monoids that are not groups).$\endgroup$
– celtschkDec 14 '18 at 8:51

$\begingroup$This is actually quite neat! There is no circular logic either, which I was initially worried about. (Also, I've edited the post to make the map clearer. Feel free to undo my edit if you wish.)$\endgroup$
– user1729Dec 14 '18 at 15:35

$\begingroup$Hi, I know this is old, but how can you multiply the left side by $x$ and right side by $y$ and not do the same for both sides? In other words, if I multiply one side by $x$, don't I have to do the same for the other side? I am not sure if I am reading that correctly.$\endgroup$
– RyanFeb 10 at 1:22

1

$\begingroup$$\color{red}{x} *x*y*\color{red}{y} \neq \color{red}{x} * y * x * \color{red}{y} $, so to be precise I should have said multiple both sides on the left by $x$ and on the right by $y$. Then applying associativity ($*$ is associative by definition of group) you get $(\color{red}{x} *x)*(y*\color{red}{y}) \neq (\color{red}{x} * y) * (x * \color{red}{y})$ and there you go. I hope it helps.$\endgroup$
– Marco BellocchiFeb 10 at 9:49

$\begingroup$Thank you for responding to an old post and for the clarification!$\endgroup$
– RyanFeb 10 at 18:43

It is easier to solve. Let be c=a○b, then c○c=e, (a○b)○(a○b)=e.
Then multiply the left and right parts of the right by b and a: (a○b)○(a○b)○b=e○b → (a○b)○(a○b○b)=e○b → (a○b)○(a○e)=e○b → (a○b)○a=b → (a○b)○a○a=b○a → (a○b)○e=b○a. the end.

$\begingroup$Your answer is nigh unintelligible. A good place to start might be with the MathJax tutorial. You also declare that your approach is "easier"---it might be worthwhile to explain why it is easier than the other answers (or, at least, to point out the flaws of the other answers).$\endgroup$
– Xander HendersonJan 14 at 0:07